Re: Cubic Equation

Re: Cubic Equation

Originally Posted by Honore

Find all positive integer roots of the equation

You can factorise it as The factor clearly cannot be a perfect square. So at least one of the squared prime factors in must be shared between the two factors on the left side. But the only possible common factor of and is 2. So we can write and for some natural numbers

Therefore which, with a bit of reorganisation and after cancelling the factor 2, can be written as This shows that the set is a Pythagorean twin triple. There are infinitely many Pythagorean twin triples, but we want one in which the middle element is a perfect square. The triples (0,1,1) and (3,4,5) satisfy that condition, leading to the values n=1 and n=7. The next triples are (20,21,29), (119,120,169) and (696,697,985), but none of those has a perfect square as the middle element. After that, the numbers increase very rapidly, as you can see from the listing at A046090 - OEIS. My guess is that there are no more perfect squares in that sequence, but I have to admit that I don't begin to see how to prove that.